circlestangent line
I am given the equation of a circle: $(x + 2)^2 + (y + 7)^2 = 25$. The radius is $5$. Center of the circle: $(-2, -7)$.
Two lines tangent to this circle pass through point $(4, -3)$, which is outside of said circle. How would I go about finding one of the equations of the lines tangent to the circle?
I haven't started calculus, so I ask for advice fitting for someone starting a topic like this.
There is a general expression derived in HCR's Web Papers for directly calculating the co-ordinates of foot of perpendicular say $(x', y')$ drawn from any point $(x_o, y_o)$ to the straight line $y=mx+c$ is given as $$(x', y')\equiv\left(\frac{x_o+m(y_o-c)}{1+m^2}, \frac{mx_o+m^2y_o+c}{1+m^2}\right)$$ Hence, the co-ordinate of the other end of diameter i.e. foot of perpendicular $(x', y')$ drawn from the point $(4, 1)$ to the line y=3x-1 are calculated as follows $$(x', y')\equiv\left(\frac{4+3(1-(-1))}{1+3^2}, \frac{3(4)+3^2(1)+(-1)}{1+3^2}\right)\equiv\left(\frac{10}{10}, \frac{20}{10} \right)\equiv\left(1, 2\right)$$ Hence, the co-ordinates of the center of the circle will be the mid-point of the line joining the end-points of the diameter $(4, 1)$ & $(1, 2)$ are calculated as follows $$\left(\frac{4+1}{2}, \frac{1+2}{2}\right)\equiv\left(\frac{5}{2}, \frac{3}{2}\right)$$
Here is a geometric version, not using a single formula. Start with the points $A$ and $B$ and a line $\ell$ through $A$ (see the figure below).
Construct the perpendicular line to $\ell$ through $A$ (the $\color{red}{\text{red}}$ line). Construct the perpendicular bisectors between $A$ and $B$ (the $\color{green}{\text{green}}$ line, the green dot is the midpoint of $A$ and $B$). The intersection of both constructed lines is the circle's center. The readius is the distance of the center to $A$.
You can translate every step into a formula to solve it numerically if necessary.
Best Answer
A useful technique for this situation which avoids using calculus goes like this:
Find the equation of a line through the point $(4,-3)$ with gradient $k$, and then use this (linear) equation to substitute into the equation of the circle to get a quadratic. The two roots of this quadratic would give the 2 points of intersection of the line with the circle, but in our case, we want to know the values of $k$ for which the quadratic has a double root (i.e. the 2 points of intersection are actually coincident). Use the standard condition ($b^2=4ac$) for a double root of a quadratic.
In your case, the equation of the line through $(4,-3)$ with gradient $k$ is $y+3 = k(x-4)$, or $y+7 = k(x-4)+4$. Substituting this into the equation of the circle gives:
$$(x+2)^2 + (k(x-4)+4))^2 = 25$$
So you need to rearrange this quadratic in $x$ and find out which vales of $k$ give double roots.